Question: $\begin{aligned} &g(x)=x^2+5x-3 \\\\ &h(y)=3(y-1)^2-5 \end{aligned}$ $(h\circ g) (-6)=$
Answer: Let's start by rewriting $(h\circ g) (-6)$ as $h(g(-6))$. When evaluating composite functions, we work our way inside out. To evaluate $h(g(-6))$, let's first evaluate $g(-6)$. Then we'll plug that result into $h$ to find our answer. Let's evaluate $g({-6})$. $\begin{aligned}g(x)&=x^2+5x-3\\\\ g({-6})&=({-6})^2+5({-6})-3~~~~~~~~~~\text{Plug in }x={-6}\\\\ &=36-30-3\\\\ &={3}\end{aligned}$ We now know that $h(g({-6}))$ is the same as $h({3})$ because $g({-6}) = {3}$. Let's evaluate $h({3})$. $\begin{aligned}h(y)&=3(y-1)^2-5\\\\ h({{3}})&=3\cdot\left(({3})-1\right)^2-5~~~~~~~~~~\text{Plug in }y={3}\\\\ &=3\cdot (2)^2-5\\\\ &=12-5\\\\ &=7\end{aligned}$ The answer: $(h\circ g)(-6) =7$